# How to compute the limit $\lim_{x\to\infty} n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}}$

I was reading this post and I got really curious about that expression. I mean, I'd like to compute the limit: $$\lim_{x\to\infty} n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}}$$

What I tried: following the advice, I took logarithms.

$$\lim_{x\to\infty} n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}} \Rightarrow \ln\left ( \lim_{x\to\infty} n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}} \right ) = n^2\cdot\ln\left ( \frac{1}{n} \right )\cdot\frac{1}{n}\cdot\sum_{k=1}^{n} \ln \left ( 1 + \frac{n}{k} \right )$$

As $\lim_{n\to\infty} \frac{1}{n}\cdot\sum_{k=1}^{n} \ln \left ( 1 + \frac{n}{k} \right ) = \int_{1}^{2} \ln(1+x) dx$, then:

$$\lim_{n\to\infty} n^2\cdot\ln\left ( \frac{1}{n} \right )\cdot\underbrace{\int_{1}^{2} \ln(1+x) dx}_{\sim 0.9 >0} = -\infty$$

Am I right?

• Where did you miss your $x$ in equation? Or there are must be another variable? – itdxer Jul 1 '15 at 17:50
• Do you mean $n$ instead of $x$ – Elaqqad Jul 1 '15 at 17:58
• Yes, you are correct. You need only evaluate $e^{\lim_{n\to \infty}\log (\cdots )}=0$ and you are done! – Mark Viola Jul 1 '15 at 20:27

Your reasoning is correct and the limit is $0$, in order to see it more clearly you can use the inequality: $$n^{-n^2} \prod_{k=1}^n (k+n)^{\frac{1}{n}}\leq \frac{\prod_{k=1}^n (2n)^{\frac{1}{n}}}{n^{n^2}}=\frac{2n}{n^{n^2}}$$ and here you can see why ...